# Estimate capacitor parameters

I am thinking of making a capacitor, just for fun/for studying purposes. And I thought of the following exercise.

In my local store, there exists rolls of polyethylene food film and aluminium foil. Theoretically, I can take a couple of both, and put them in a "sandwitch" manner, while shift layers to the left/right (to make poles for current collection). Then just roll. Okay, english is not my native, so I draw an illustration:

What I want to know - is how to estimate resulting capacitor parameters?

Given:

• roll length L
• layer width w
• aluminum thickness a
• plastic thickness p
• bobbin diameter b

I want to know:

• breakdown voltage V = ?
• capacitance C = ?
• short circuit current I = ? (at full charge, edges are instantly shorted with zero-resistance ideal superconductor)

Also, consider everything ideal:

• no air gaps between layers
• ideal winding, no tensions/etc
• normal conditions density, materials are non-compressable.

PS. This is theoretocal question, for sake of study and understanding relationship between parameters. If you have spare weekend evening. Many years ago I havent learned it at school (neither at institution). So today it would be good if I could learn something from it. Thanks.

• Great that you are doing some practical experiments.Clue. In your given section you will also need the relative permittivity of the plastic and its breakdown voltage you probably don't need the bobbin diameter. – RoyC Apr 14 at 16:54
• Use the formula C = Eo * Er * Area/Distance. I made one such parallel-plate capacitor (described by the formula) while reading the ARRL Ham Radio books, with area about 2 square feet; I estimated 0.017uF, again using waxed-paper. Roll it tight, to exclude most of the air. – analogsystemsrf Apr 14 at 17:08
• @RoyC Sorry, In the food store they just dont know permittivity of plastic film, neither its chemical formula. Its just usual food film, thay you may use to pack cheese :^) As for diameter, I thought that it may introduce something, because in inductor's case it affects number of turns you may made, so for infinite inductance you want to shrink inductor to a single "mass" point, with "infinite number of turns" inside it – xakepp35 Apr 14 at 17:23
• Wrapping film is generally LPDE, 8 ­– 12.5μm thick. wikipedia plastic wrap – CSM Apr 14 at 19:25
• Please try and report back with the results using a whole roll. – pipe Apr 14 at 21:54

Capacitance is given by:

This is the formula applicable for free space or vacuum. If you are using a dielectric material between the plates, you multiply the resultant capacitance with the relative permittivity of the dielectric material. For poly-ethylene, relative permitivitty is 2.25.

For simplicity, lets start with flat sheets of aluminium separated by a polyethylene film:

Assuming polyethylene thickness = 15 microns And dimensions to be 20 cm x 50 cm

Your capacitance will be C = 2.25*8.85418782 × 10^-12*0.2*0.5/0.000015 F = 0.133 uF

Now, if you intend to roll the flat sheets on a bobbin, things will change. Consider the image below:

Now, all of a sudden, both sides of aluminium foil start acting as capacitors. Black is one plate and blue is another plate of aluminium. White space is dielectric.

This should approximately double the capacitance.

Now coming to your questions, break down voltage appears to be very different across different sources:

Capacitance calculated above.

Ideally there should be infinite current at zero resistance discharge.

• As for discharge: The foil is a rectangular conductor, in crosssection. its area = thickness * length. aluminum conductance is also limited. so if you imagine a shour circuit (turn your bobbin into torus, to connect its both edges) there will be non infinite depolarisation current. – xakepp35 Apr 14 at 17:28
• Could I just use ohm's law? (where V=breakdown voltage, and R=resistance of rectangular pure aluminum conductor, and its length is equal to roll width?) But no, charge is stored uniformly along roll width. So that is also required to take into account – xakepp35 Apr 14 at 17:29
• No the leakage is due to contaminants in parts per billion and also the above GV/m or kV/mm values are under pristine ideal conditions with no safety factor unlike XLPE HVAC cables which are de-rated. – Tony Stewart Sunnyskyguy EE75 Apr 14 at 17:44
• Practically, there will be a finite but large current because any charge movement will happen through the aluminium. There might be some analysis to give a somewhat approximate answer to your question but sadly I am not qualified enough to answer that. – Whiskeyjack Apr 14 at 17:44
• don't forget $\epsilon _R$ – Tony Stewart Sunnyskyguy EE75 Apr 14 at 17:50

$$\C= \epsilon _o \cdot \epsilon _R\dfrac{A}{h}\$$
where $$\D_k=\epsilon _R = \$$ = "relative permittivity" : 2 to 5 for most plastic, FR4 = 4.2 @ 10 MHz

https://www.digikey.com/eewiki/display/Motley/Film+Capacitors

film capacitor sub-type in which the electrodes are built up on a supporting substrate in a very thin (10’s of nanometers) layer, usually through a vacuum deposition process.

If you have calipers and fold a large sheet of each material 10 times in half you have 1024 x the original thickness. Then squeeze the air out outside your foil.

Leave extra plastic on the outside and then roll around something like a pencil and insert some foil tabs for electrodes. If you puncture the film , you won't get the full Dielectric Breakdown Voltage rating for PE (polyethylene) plastic.

# BDV vs Material ( smooth flat electrodes)

my estimates from experience

3.0 kV/mm very clean Air
1 kV/mm dusty humid air

5   kV/mm clean PU and smooth foil
10  kV/mm processed PU in capacitors
20  kV/mm cross-link PU (XLPU) processed in clean rooms with 1 GV

25 kV/mm min Transformer Oil as shipped
75 kV/mm min Transformer Oil process by machine: heat, vacuum, demoist, and HEPA oil filter


Can you get 10nF? 0.1uF?? with 5kV/mm of dielectric thickness? ( 5V/um )

Good luck. Walk around with neoprene shoes on a nylon carpet to charge up Although PU caps have very low resistance on the electrodes which controls the ESR, you won't get that with alum foil. They etch it with acid to increase the nano-surface roughness.

• They usually print thickness, for example 8um... So, if I understand correctly, with typical PET film of 8um thickness I may get ~40V breakdown voltage? – xakepp35 Apr 14 at 17:31
• Yes 40V is safe. But it's non-self-healing. So if zapped its a short. If you ramp up DC and monitor with a SW radio for lightning pulses tick tick just before the arc, with current limiting you might be able to hear the partial discharges before it zaps the smallest gap. I did this for a power transformer and that works well a pulse of 1ms will be loud on the radio but I would expect 100ns pulse to be attenuated 80dB in a 10kBW @ >0.5Mhz AM or SW but may not hear it in acoustics go tick. It might survive 100Vdc if everything is clean and nF value works out correct to prove gap with Dk=3 – Tony Stewart Sunnyskyguy EE75 Apr 14 at 19:09
• e-caps have an ESRC = 200us and low ESR e-caps down to 1 to 10 us ( some expensive ones less) while film caps have ESRC value in the x to xx ns range, while high performance film caps for Xray are better and I would expect worse here. – Tony Stewart Sunnyskyguy EE75 Apr 14 at 19:13
• Also the ESL or inductance of a 2mm x 2mm square is the same as 1m x 1m so it depends on L/W ratio. and to a small extent thickness of conductor. So a tall cylinder cap has less ESL and less ESR than a short one of same voltage and rating – Tony Stewart Sunnyskyguy EE75 Apr 14 at 19:22
• Are you missing free space permittivity in your first equation? – RoyC Apr 23 at 8:17